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Contents
CHAPTER 14 Multiple Integrals
14.1 Double Integrals
14.2 Changing to Better Coordinates
14.3 Triple Integrals
14.4 Cylindrical and Spherical Coordinates
CHAPTER 15 Vector Calculus
15.1 Vector Fields
15.2 Line Integrals
15.3 Green's Theorem
15.4 Surface Integrals
15.5 The Divergence Theorem
15.6 Stokes' Theorem and the Curl of F
CHAPTER 16 Mathematics after Calculus
16.1 Linear Algebra
16.2 Differential Equations
16.3 Discrete Mathematics
Study Guide For Chapter 1
Answers to OddNumbered Problems
Index
Table of Integrals
CHAPTER 14
Multiple Integrals
14.1 Double Integrals 4
This chapter shows how to integrate functions of two or more variables. First, a
double integral is defined as the limit of sums. Second, we find a fast way to compute
it. The key idea is to replace a double integral by two ordinary "single" integrals.
The double integral Sf f(x, y)dy dx starts with 1f(x, y)dy. For each fixed x we integ
rate with respect to y. The answer depends on x. Now integrate again, this time with
respect to x. The limits of integration need care and attention! Frequently those limits
on y and x are the hardest part.
Why bother with sums and limits in the first place? Two reasons. There has to be
a definition and a computation to fall back on, when the single integrals are difficult
or impossible. And alsothis we emphasizemultiple integrals represent more than
area and volume. Those words and the pictures that go with them are the easiest to
understand. You can almost see the volume as a "sum of slices" or a "double sum of
thin sticks." The true applications are mostly to other things, but the central idea is
Add up small pieces and take limits.
always the same:
We begin with the area of R and the volume of by double integrals.
A LIMIT OF SUMS
The graph of z =f(x, y) is a curved surface above the xy plane. At the point (x, y) in
the plane, the height of the surface is z. (The surface is above the xy plane only when
z is positive. Volumes below the plane come with minus signs, like areas below the
x axis.) We begin by choosing a positive functionfor example z = 1+ x2 + y2.
The base of our solid is a region R in the xy plane. That region will be chopped
into small rectangles (sides Ax and Ay). When R itself is the rectangle 0d x < 1,
0< y < 2, the small pieces fit perfectly. For a triangle or a circle, the rectangles miss
part of R. But they do fit in the limit, and any region with a piecewise smooth
boundary will be acceptable.
Question What is the volume above R and below the graph of z =Ax, y)?
Answer It is a double integralthe integral of f(x, y) over R. To reach it we begin
with a sum, as suggested by Figure 14.1.
14 Multiple Integrals
area AA
Fig. 14.1 Base R cut into small pieces AA. Solid V cut into thin sticks AV = z A A.
For single integrals, the interval [a, b] is divided into short pieces of length Ax.
For double integrals, R is divided into small rectangles of area AA = (Ax)(Ay). Above
the ith rectangle is a "thin stick" with small volume. That volume is the base area
AA times the height above itexcept that this height z =f(x, y) varies from point to
point. Therefore we select a point (xi, y,) in the ith rectangle, and compute the volume
from the height above that point:
volume of one stick =f(xi, yi)AA volume of all sticks = 1f(xi, yi)AA.
This is the crucial step for any integralto see it as a sum of small pieces.
Now take limits: Ax + 0 and Ay + 0. The height z =f(x, y) is nearly constant over
each rectangle. (We assume that f is a continuous function.) The sum approaches a
limit, which depends only on the base R and the surface above it. The limit is the
volume of the solid, and it is the double integral of f(x, y) over R:
f(x, y) dA = lim 1f(xi, yi)AA.
J JR Ax t 0
Ay+O
To repeat: The limit is the same for all choices of the rectangles and the points (xi, yi).
The rectangles will not fit exactly into R, if that base area is curved. The heights are
not exact, if the surface z =f(x, y) is also curved. But the errors on the sides and top,
where the pieces don't fit and the heights are wrong, approach zero. Those errors are
the volume of the "icing" around the solid, which gets thinner as Ax + 0 and Ay + 0.
A careful proof takes more space than we are willing to give. But the properties of
the integral need and deserve attention:
jj(f + g)dA = jj f d~ + jjg dA
1. Linearity:
2. Constant comes outside: jj cf(x, y)dA = c jj f(x, y)dA
3. R splits into S and T(not overlapping): ]jf d~ = jj fd~+ jj f d~.
R S T
In 1 the volume under f + g has two parts. The "thin sticks" of height f + g split into
thin sticks under f and under g. In 2 the whole volume is stretched upward by c. In
3 the volumes are side by side. As with single integrals, these properties help in
computations.
By writing dA, we allow shapes other than rectangles. Polar coordinates have an
extra factor r in dA = r dr do. By writing dx dy, we choose rectangular coordinates
and prepare for the splitting that comes now.
14.1 Double Integrals
SPLITTING A DOUBLE INTEGRAL INTO TWO SINGLE INTEGRALS
The double integral jjf(x, y)dy dx will now be reduced to single integrals in y and
then x. (Or vice versa. Our first integral could equally well be jf(x, y)dx.) Chapter 8
described the same idea for solids of revolution. First came the area of a slice, which
is a single integral. Then came a second integral to add up the slices. For solids
formed by revolving a curve, all slices are circular disksnow we expect other shapes.
Figurle 14.2 shows a slice of area A(x). It cuts through the solid at a fixed value of
x. The cut starts at y =c on one side of R, and ends at y =d on the other side. This
particular example goes from y =0 to y =2 (R is a rectangle). The area of a slice is
the y integral of f(x, y). Remember that x is fixed and y goes from c to d:
A(x) =area of slice = f(x, y)dy (the answer is a function of x).
Icd
EXAMPLE I A= (1 +x2 +y2)dy=
This is the reverse of a partial derivative! The integral of x2dy, with x constant, is
x~~.This "partial integral" is actually called an inner integral. After substituting the
limits y =2 and y =0 and subtracting, we have the area A(x) =2 +2x2 +:. Now the
outer integral adds slices to find the volume j A(x) dx. The answer is a number:
rc
X
Fig. 14.2 A slice of V at a fixed x has area A(x)= f(x, y)dy.
To complete this example, check the volume when the x integral comes first:
y=2 8 8 16
outer integral =~y~o(~+y2)dy=[~y+~y3] y=O =+=3 3 3 '
The fact that double integrals can be split into single integrals is Fubini's Theorem.
I 14A If f(x, y) is continuous on the rectangle R, then I
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